[tex]dane:\\s_1 = 50 \ km\\t_1 = 3 \ h\\s_2 = 90 \ km\\t_2 = 2 \ h\\szukane:\\v_{sr} = ?\\\\v_{sr} = \frac{s_{c}}{t_{c}}\\\\s_{c} = s_1 + s_2 = 50 \ km + 90 \ km = 140 \ km\\t_{c} = t_1 + t_2 = 3 \ h + 2 h\\\\v_{sr} = \frac{140 \ km}{5 \ h} = 28\frac{km}{h}[/tex]
[tex]v_1 = \frac{s_1}{t_1} = \frac{50 \ km}{3 \ h} = 16,(6)\frac{km}{h}\approx 16,67\frac{km}{h}\\\\v_2 = \frac{s_2}{t_2} = \frac{90 \ km}{2 \ h} = 45\frac{km}{h}[/tex]
